%!TEX root = thesis.tex
\chapter{Del Pezzo Surfaces of Degree $\le 4$}\label{chap:DelPezzo}

We are finally in a position to prove Theorem \ref{thm:ed2classification}.  It remains only to show that groups with versal actions on del Pezzo surface of degree $\le 4$ have already been seen acting versally on the surfaces of Theorem \ref{thm:4surfacesClassification}.  Indeed, the main theorem is an immediate consequence of the following:

\newtheorem*{ed2to4surfacesHack}{Theorem \ref{thm:ed2to4surfaces}}
\begin{ed2to4surfacesHack}
If $G$ is a finite group of essential dimension $2$ then $G$ has a versal action on $\bbP^2$, $\bbP^1 \times \bbP^1$, $DP_6$, or $DP_5$.
\end{ed2to4surfacesHack}

\begin{proof}
All groups $G$ of essential dimension $2$ have versal actions on minimal rational $G$-surfaces by Proposition \ref{prop:ratMinModelStrategy}.  Thus, it suffices to prove that, for any minimal rational versal $G$-surface $X$, there exists a versal action on one of the $4$ surfaces listed above.  Recall that any minimal rational $G$-surface $X$ is a del Pezzo surface or has a conic bundle structure by Theorem \ref{thm:EMI}.

Theorem \ref{thm:conicVersal} proves the theorem for surfaces with a conic bundle structure.  We recall from \cite[Section 6]{DolgachevIskovskikh2006Finite-subgroup} that the only minimal rational $G$-surfaces of degree $\ge 5$ are precisely those listed in the statement of the theorem.  Thus it suffices to consider degrees $\le 4$.  In the following $X$ is a del Pezzo surface with a versal $G$-action.

\smallskip \noindent
{\bf Case degree $4$:}
The minimal groups of automorphisms of del Pezzo surfaces $X$ of degree $4$ are listed in \cite[Theorem 6.9]{DolgachevIskovskikh2006Finite-subgroup}.  We know that $G$ must be from this list and that $\ed(G) \le 2$.  If $G$ is abelian or a $2$-group then it acts versally on $\bbP^2$.  All remaining groups have abelian subgroups with ranks $\ge 3$ (note that $C_2 \times A_4$ contains $C_2^3$); thus they cannot be versal.

An alternative proof that does not rely directly on \cite[Theorem 6.9]{DolgachevIskovskikh2006Finite-subgroup} can be found in Section \ref{sec:altProofs}.

\smallskip \noindent
{\bf Case degree $3$:}
The minimal groups of automorphisms of del Pezzo surfaces $X$ of degree $3$ are listed in \cite[Theorem 6.14]{DolgachevIskovskikh2006Finite-subgroup}.  It suffices to consider $G$ from this list.

All groups with non-trivial centres and essential dimension $\le 2$ have versal actions on $\bbP^2$ by Proposition \ref{prop:centreMeansP2}.  Thus we may assume $G$ has a trivial centre.  In particular, we may eliminate all abelian groups from the list.  Next, we may eliminate all groups with abelian subgroups of rank $\ge 3$ since they cannot be versal by Proposition \ref{prop:edResults}(\ref{prop:edResults:abelian}).  Similarly, we eliminate $G$ containing a non-abelian $3$-subgroup by Proposition \ref{prop:3groupsAbelian}.  Also, if $G$ is a subgroup of $S_5$ then $G$ has a versal action on $DP_5$ by the proof of Theorem \ref{thm:4surfacesClassification}.

All that remains to consider are $G$ of the form $C_3^2 \rtimes C_2$ and $C_3^2 \rtimes C_2^2$.  It suffices to consider $G \simeq C_3^2 \rtimes C_2^2$.  We may view this group as a representation of $C_2^2$ on the vector space $\bbF_3^2$.  Since the centre is trivial, we may assume the representation is faithful.  The representation is diagonalisable, so $G$ is isomorphic to $S_3 \times S_3$.  This group has a versal action on $\bbP^1 \times \bbP^1$ by the proof of Theorem \ref{thm:4surfacesClassification}.

An alternative proof can be found in Section \ref{sec:altProofs}.

\smallskip \noindent
{\bf Case degree $2$:}
We have a finite $G$-equivariant morphism of degree $2$ to $\bbP^2$ (see \cite[Section 6.6]{DolgachevIskovskikh2006Finite-subgroup}).  If the induced action of $G$ on $\bbP^2$ is faithful then we are done.  Otherwise, the group $G$ contains a central involution (a \emph{Geiser involution}).  Any such group has a non-trivial centre and sits inside $\GL_2(\bbC)$ by Corollary \ref{prop:centreMeansP2}.  

\smallskip \noindent
{\bf Case degree $1$:}
This case proceeds the same way as degree $2$ via the \emph{Bertini involution}.  The only difference is that the finite morphism of degree $2$ maps onto a singular quadric cone in $\bbP^3$ (\cite[Section 6.7]{DolgachevIskovskikh2006Finite-subgroup}).  The automorphism group of a singular quadric cone is the same as the minimal ruled surface $\bfF_2$ (see \cite[Example V.2.11.4]{Hartshorne1977Algebraic-geome}).  Any versal action on such a surface must also act versally on $\bbP^2$ or $\bbP^1 \times \bbP^1$ (see Lemma \ref{lem:conicRuled} above).
\end{proof}

\section{Alternative Proofs}\label{sec:altProofs}

Recall that the degree $3$ and $4$ cases of Theorem \ref{thm:ed2to4surfaces} above used Dolgachev and Iskovskikh's classification \cite{DolgachevIskovskikh2006Finite-subgroup} in an essential way.  The careful reader will find that the proofs in the classification are quite involved.  To ease the reader's burden, we supply alternative proofs for these two cases.  Our context is considerably less ambitious than theirs, so we have the luxury of a simplified argument.

\subsection{Del Pezzo Surfaces of Degree $4$}

\begin{thm}\label{thm:DP4Versal}
Suppose $G$ is a finite group with a versal action on a del Pezzo Surface $X$ of degree $4$.  Then $G$ has a versal action on $\bbP^2$ or $DP_5$.
\end{thm}

\begin{proof}
Recall from \cite{DolgachevIskovskikh2006Finite-subgroup} or \cite{Manin1986Cubic-forms} that groups acting on a del Pezzo surface of degree $4$ must sit inside the Weyl group $W(D_5) \simeq C_2^4 \rtimes S_5$.  We see that any group with a versal action on such a surface must also act versally on $\bbP^2$ or $DP_5$ via Lemma \ref{lem:ed2inWD5} below.
\end{proof}

\begin{lem}\label{lem:ed2inWD5}
If $G$ is a subgroup of $W(D_5) \simeq C_2^4 \rtimes S_5$ of essential dimension $2$ then $G \subset \GL_2(\bbC)$ or $G \subset S_5$.
\end{lem}

\begin{proof}
Let $K = G \cap C_2^4$.  Clearly, $K$ must have rank $\le 2$.  If $K$ has rank $0$ then $G$ embeds into $S_5$.  If $K$ has rank $1$ it must be central in $G$, so $G$ has a non-trivial centre and embeds in $\GL_2(\bbC)$ by Proposition \ref{prop:centreMeansP2}.  It remains to consider $K$ of rank $2$, where $K \simeq C_2 \times C_2$.

If $5$ divides $|G|$ then $G$ contains an element $\sigma$ of order $5$.  The action of $\sigma$ on $K$ must be trivial since $K$ has fewer than $5$ elements.  The origin of $K$ is the only element that can be invariant under the $\sigma$ action.  Thus $G$ has no elements of order $5$.

Note that the automorphism group of $K$ is $S_3$.  Since $K$ is a normal subgroup of $G$ there exists a map $G \to \Aut(K) \simeq S_3$.  If an element of $K$ is fixed by the $G$-action then $G$ has a non-trivial centre and $G \subset \GL_2(\bbC)$.  Thus it suffices to assume the image of $G \to \Aut(K)$ must contain an element of order $3$.  Since $S_5$ has no elements of order $9$, we may assume there exists $\sigma \in G$ of order $3$ which acts transitively on the non-trivial elements of $K$.

It is useful to view $V = C_2^4$ as a vector subspace $a_1+a_2+a_3+a_4+a_5=0$ for $(a_1,a_2,a_3,a_4,a_5) \subset \bbF_2^5$ with $S_5$ permuting basis elements in the obvious way.  We may choose our coordinates so that $\sigma \mapsto (123)$ under the map $G \to G/K \subset S_5$.  Let $x = (a_1,a_2,a_3,a_4,a_5) \in K \setminus \{0\}$, we see that
\begin{align*}
y &= x + x^\sigma + x^{\sigma^2} \\
&= (a_1+a_2+a_3,a_1+a_2+a_3,a_1+a_2+a_3,a_4,a_5).
\end{align*}
Note that $y$ must be in $K$.

If $y=0$ then $a_4=a_5=0$ and $a_1+a_2+a_3=0$.  The only non-trivial $K$ satisfying these conditions is $\langle (1,1,0,0,0), (0,1,1,0,0) \rangle$.  The stabiliser of this group in $S_5$ is of the form $S_3 \times C_2$ with $S_3$ permuting the first $3$ basis vectors and $C_2$ swapping the last $2$.  The action of $S_3$ on $\langle (1,1,0), (0,1,1) \rangle \subset \bbF_2^3$ yields a group isomorphic to $S_4$.  We see then that $G$ is a subgroup of $S_4 \times C_2$.  If the restriction to $G$ of the projection map $S_4 \times C_2 \to S_4$ is an isomorphism then $G \subset S_4 \subset S_5$; otherwise $G$ has a non-trivial centre and is in $\GL_2(\bbC)$.

If $y\ne 0$ then $y=x^g$ for some $g \in \langle \sigma \rangle$.  Thus $x+x^g+x^{g^2}=x^g$ and then $x+x^{g^2}=0$.  Thus $a_1=a_2=a_3$.  There are only $3$ such elements in $V$: $(0,0,0,1,1)$, $(1,1,1,1,0)$ and $(1,1,1,0,1)$.  All of these elements are fixed by $\sigma$ so this case does not occur.
\end{proof}

\subsection{Del Pezzo Surfaces of Degree $3$}

The finite groups acting on del Pezzo surfaces of degree $3$ were classified by Segre \cite{Segre1942The-Non-singula}.  Unfortunately, Segre's classification had errors as pointed out by more recent classifications by Dolgachev and Iskovskikh \cite{DolgachevIskovskikh2006Finite-subgroup}, and by Hosoh \cite{Hosoh1997Automorphism-grou}.  However, even the newer classifications are long and technical (this is probably inevitable).  Given the checkered history of this problem, it seems wise to have an independent argument in our more modest context.

In addition, many of our intermediate results are stronger than the classification alone.  These may be useful for extending the results to more general fields.

\begin{thm}\label{thm:DP3Versal}
Suppose $G$ is a finite group with a versal action on a del Pezzo surface $X$ of degree $3$.  Then $G$ has a versal action on $\bbP^2$, $\bbP^1 \times \bbP^1$ or $DP_5$.
\end{thm}

\begin{proof}

We recall some basic facts from \cite[Section 6.5]{DolgachevIskovskikh2006Finite-subgroup}.

Any del Pezzo surface $X$ of degree $3$ has a $G$-equivariant embedding $X \hookrightarrow \bbP^3$ as a non-singular cubic hypersurface obtained via the anti-canonical sheaf $-K_X$.  As in \cite[\S 4]{Iskovskih1979Minimal-models-}, we have a representation $\rho$ of $G$ acting on the $4$-dimensional space $V=\Gamma(X,-K_X) \simeq \bbC^4$ which lifts the action on $\bbP^3$.  Let $F \in \bbC[x_1,x_2,x_3,x_4] = \bbC[V]$ be the homogeneous form of degree $3$ corresponding to $X$ in $\bbP^3$.

The non-singular condition puts some restrictions on the possibilities for $F$:

\begin{lem}\label{lem:minMonomialExp}
For every $x_i$, there must be a monomial in $F$ such that $x_i$ has exponent $\ge 2$.
\end{lem}

\begin{proof}
This is proved in \cite{DolgachevIskovskikh2006Finite-subgroup}; we include it here for the sake of completeness.  Assume the contrary, that $F$ does not have a monomial where $x_i$ occurs with exponent $\ge 2$.  Without loss of generality, we may assume it is $x_1$.  The polynomial $F$ can be written in the form $x_1G(x_2,x_3,x_4) + H(x_2,x_3,x_4)$ where $G$ and $H$ are forms of order $2$ and $3$ respectively.  The point $(1:0:0:0) \in X$ is singular; a contradiction.
\end{proof}

For the rest of the proof we assume that $G$ has a versal action on $X$.  The remainder of the proof is structured as follows.  First, we show that, if $\rho$ is reducible, $G$ must act versally on $\bbP^2$ or $\bbP^1 \times \bbP^1$ (Lemma \ref{lem:cubicReducible}).  Next, we prove that if $G$ has an element of order $5$ it must act versally on $DP_5$ (Lemma \ref{lem:cubicNo5}).  Of the remaining groups, we prove that there are no irreducible monomial versal actions (Lemma \ref{lem:cubicMonoVersal}).  Finally, we eliminate all irreducible actions (Lemma \ref{lem:cubicIrred}).

\begin{lem}\label{lem:cubicReducible}
If $\rho$ is reducible then $G$ has a versal action on $\bbP^2$ or $\bbP^1 \times \bbP^1$.
\end{lem}

\begin{proof}
If $G$ has a non-trivial centre then $G \subset \GL_2(\bbC)$ and we have a versal action on $\bbP^2$ by Proposition \ref{prop:centreMeansP2}.  Thus we may assume $Z(G)=1$.  Let $\rho = \sigma_1 \oplus \sigma_2$ be a decomposition into subrepresentations where $\dim(\sigma_1) \ge \dim(\sigma_2)$ and $\sigma_2$ is irreducible.

Consider the case where $\dim(\sigma_2)=1$.  Suppose there exists an element $g \in G$ such that $\sigma_1(g)=1$ but $\sigma_2(g)\ne 1$.  Any such $g$ would be in the centre of $G$, so no such elements exist.  This means that $\sigma_1$ is faithful.  We have a $3$ dimensional faithful representation of $G$ with a trivial centre.  Thus, $G$ acts versally on $\bbP^2$ by Theorem \ref{thm:VersalCoxSplit}.

We are left with the case where $\sigma_1$ and $\sigma_2$ are both irreducible of dimension $2$.  Since the centre of $G$ is trivial, $G$ descends to a versal action on $\bbP^1 \times \bbP^1$ by Theorem \ref{thm:VersalCoxSplit}.
\end{proof}

Recall that we have an embedding $G \hookrightarrow W(E_6)$ where $W(E_6)$ is the Weyl group associated to the root system $E_6$.  Thus $|G|$ divides $|W(E_6)|=51840=2^7\cdot 3^4\cdot 5$.  We handle the case where $G$ has an element of order $5$:

\begin{lem}\label{lem:cubicNo5}
If $G$ has an element of order $5$ then $G$ has a versal action on $\bbP^2$ or $DP_5$.
\end{lem}

\begin{proof}
Any such group $G$ must act on the surface of type II in the table from \cite[Section 6.5]{DolgachevIskovskikh2006Finite-subgroup}.  The automorphism group of this surface is $S_5$.  So, in fact we have the stronger result that $G \subset S_5$ regardless of whether $G$ is versal.  Since the proof in \cite{DolgachevIskovskikh2006Finite-subgroup} involves complicated calculations, we sketch an alternate proof here using the fact that $G$ is versal.

The group $G$ must be isomorphic to a subgroup of $W(E_6)$.  The maximal subgroups of $W(E_6)$ with order divisible by $5$ are $C_2^4 \rtimes S_5$ and $S_6 \times C_2$.  By Theorem \ref{lem:ed2inWD5}, $G \subset C_2^4 \rtimes S_5$ implies that $G \subset S_5$ or $G$ has a non-trivial centre.  In the case where $G \subset S_6 \times C_2$, if the projection $S_6 \times C_2 \to S_6$ is not injective when restricted to $G$ then $G$ has a non-trivial centre (and so acts versally on $\bbP^2$).  Thus, it remains to consider $G \subset S_6$.

The group $S_6$ has essential dimension $3$, so $G$ cannot be the whole group.  All maximal subgroups of $S_6$ of order divisible by $5$ are isomorphic to $A_6$ or $S_5$.  All faithful representations of $A_6$ have dimension $\ge 5$ so this group does not occur either.  All maximal subgroups of $A_6$ of order divisible by $5$ are isomorphic to $A_5$.  Thus $G \subset S_5$ and has a versal action on $DP_5$.
\end{proof}

There exists a versal cubic surface realising an $S_5$-action appearing in the lemma above (see \cite{Kraft2006A_result_of_Her}).

All remaining cases have $|G|$ coprime to $5$ and $\rho$ irreducible.  We will show that none of the remaining cases can occur.  We consider the case where $\rho$ is an irreducible monomial action:

\begin{lem}\label{lem:cubicMonoVersal}
If $|G|$ is coprime to $5$ then $\rho$ cannot be irreducible monomial.
\end{lem}

\begin{proof}
Since $\rho$ is irreducible, all central elements are scalar matrices and act trivially on $X$.  Thus, we may assume $G$ has a trivial centre.

The monomial representation on $V$ gives us an extension $1 \to K \to G \to H \to 1$ where $K$ is an abelian group and $H$ is a subgroup of $S_4$.  We may assume $x_1, \ldots, x_4$ is a monomial basis for $V^*$ on which $K$ acts by diagonal matrices and $H$ acts by ``twisted'' permutations (see Section \ref{sec:toricVersal:monomial}).  We require that $H$ is a transitive subgroup of $S_4$ or else $\rho$ is reducible.  Also, we require that $K$ is non-trivial or else $\rho$ is reducible (since all irreducible representations of $S_4$ are of dimension $\le 3$).  Let $\chi_i$ be the character corresponding to the action of $K$ on $x_i$ for each $i = 1, 2, 3, 4$.

\smallskip \noindent
{\bf Case 1: $\chi_i$ are not distinct.}

Without loss of generality, we may assume $\chi_1=\chi_2$.  By transitivity of $H$, this means that $\chi_3=\chi_4$.  The group $H$ permutes the subsets $\{1,2\}$ and $\{3,4\}$.  The possible $H$ are all subgroups of the following group isomorphic to $D_8$:
\[ \langle (1\ 2), (1\ 3\ 2\ 4) \rangle \ . \]
Note that $H$ is a $2$-group.  Let $P$ be a $2$-Sylow subgroup of $G$ containing $H$.  We find $G=PK$.  Recall that any normal subgroup of a $2$-group must intersect the centre, so if $|K|$ is even then $K$ must contain an element $s$ in the centre of $P$.  Thus $s$ is in the centre of $G$; a contradiction.

We may assume $|K|$ is odd.  Since $H$ is transitive, $h = (1\ 2)(3\ 4)$ is an element of $H$.  Note that $h$ is central in $H$ and acts trivially on $K$.  Since $|K|$ and $|H|$ are coprime, $G$ splits as $K \rtimes H$.  Thus $h$ lifts to a central involution in $G$; a contradiction.

\smallskip \noindent
{\bf Case 2: all $\chi_i$ are distinct.}

Suppose $F$ contains a monomial of the form $x_i^3$.  $F$ must contain all monomials of the form $x_i^3$ by transitivity of $H$.  Since all $\chi_i$ are distinct, the only fixed points of $K$ acting on $\bbP(V)$ are
\[ \{ (1:0:0:0), (0:1:0:0), (0:0:1:0), (0:0:0:1) \}.\]
Since $x_1^3$ is the only degree $3$ monomial non-zero on $(1:0:0:0)$, this point is not on the surface $X$.  Similarly, none of the other points are on $X$.  Thus we have an abelian subgroup $K$ without fixed points on $X$.  This contradicts Proposition \ref{prop:abelSubgroups}.  We may assume $F$ contains no monomials of the form $x_i^3$.

From Lemma \ref{lem:minMonomialExp}, for every $i \in 1, 2, 3, 4$ the polynomial $F$ must contain a monomial where $x_i$ has power $\ge 2$.  Thus, $F$ contains a monomial of the form $x_i^2x_j$ for $i \ne j$.  If $F$ also contains the monomial $x_ix_j^2$ then $2\chi_i+\chi_j = \chi_i+2\chi_j$ and we find $\chi_i=\chi_j$, a contradiction.  In particular, no permutation of $H$ contains $(i\ j)$ in its cycle decomposition.  If $F$ contains a monomial $x_i^2x_k$ for $k \ne j$ then $2\chi_i+\chi_j = 2\chi_i+\chi_k$ and we find $\chi_j=\chi_k$, a contradiction.  Thus, we may assume there is exactly one monomial of the form $x_i^2x_j$ for any particular $i$.  In particular, any element of $H$ leaves $i$ fixed if and only if it leaves $j$ fixed.

Without loss of generality, we may assume that $x_1^2x_2$ is in $F$.  Since $H$ is transitive, we may assume there is an involution $h \in H$ which does not fix $\{1\}$.  This element cannot contain $(1\ 2)$ in its cycle decomposition and if it moves $\{1\}$ it must move $\{2\}$.  The only possibilities are $h=(1\ 3)(2\ 4)$ or $h=(1\ 4)(2\ 3)$.  Without loss of generality, we take $h=(1\ 3)(2\ 4)$.

There must also be $x_2^2x_j$, $x_3^2x_k$, and $x_4^2x_l$ in $F$ for some $j,k,l$.  With the considerations above, the only possible monomials are $x_2^2x_3$, $x_3^2x_4$, and $x_4^2x_1$.

Since all of these monomials are present:
\[2\chi_1+\chi_2=2\chi_2+\chi_3=2\chi_3+\chi_4=2\chi_4+\chi_1.\]
Combining adjacent equalities we have $\chi_2 = 2\chi_1-\chi_3$, $\chi_3 = 2\chi_2-\chi_4$ and $\chi_4 = 2\chi_3-\chi_1$.  Substituting into the second equation we find: $5\chi_3 = 5\chi_1$.  Since $K$ is an abelian group of order coprime to $5$ this means $\chi_1=\chi_3$.  This contradicts our assumption that the $\chi_i$ are distinct.
\end{proof}

The remaining cases can all be reduced to the monomial case:

\begin{lem}\label{lem:cubicIrred}
If $|G|$ is coprime to $5$ then $\rho$ cannot be irreducible.
\end{lem}

\begin{proof}
We will show that any irreducible $\rho$ must be also be monomial and so is impossible by Lemma \ref{lem:cubicMonoVersal} above.  Recall that $\rho$ cannot contain any scalar matrices.  Since $\rho$ is irreducible, this means $G$ has a trivial centre.  

It suffices to prove the theorem for $G$ where the restricted representations of all subgroups are reducible.  Since we are assuming $|G| = 2^a3^b$, we see that $G$ is solvable by Burnside's theorem.  Thus, we may assume $G$ has a non-trivial normal subgroup $H$ whose restricted representation $\rho|_H$ is reducible.  From \cite[Proposition 24]{Serre1977Linear-represen}, $\rho|_H$ is isotypic or there exists a group $M$ containing $H$, and an irreducible representation $\sigma$ of $M$ such that $\rho$ is induced by $\sigma$.

\smallskip \noindent
{\bf Case 1: $\rho|_H$ is isotypic.}

If $\rho|_H$ is a direct sum of linear characters then all its elements are scalar matrices.  Since $G$ has no scalar matrices, $H$ must be trivial: a contradiction.  Thus, $\rho|_H$ is a direct sum of $2$ copies of a $2$-dimensional irreducible representation $\sigma$.

If $\rho|_H$ has a non-trivial centre then $G$ has a non-trivial centre.  All $2$-groups have non-trivial centres, so $H$ is not a $2$-group.  There exists an element $h \in H$ of order $3$. Since $h$ is not central, $\sigma(h)$ has distinct eigenvalues.  Thus, $\rho(h)$ has $2$ eigenvalues, each with multiplicity $2$.  We may apply Lemma \ref{lem:A2toFermat} below to see that $\rho$ is monomial.

\smallskip \noindent
{\bf Case 2: $G$ is induced from $(M,\sigma)$.}

We either have $\dim(\sigma)=1$ or $\dim(\sigma)=2$.  If $\dim(\sigma)=1$ then $\rho$ is monomial.  The only remaining case is where $\dim(\sigma)=2$.

We see that $M$ has index $2$ in $G$, so $M$ is a normal subgroup of $G$.  We may write $\rho|_M=\sigma \oplus \sigma'$ for a $2$-dimensional representation $\sigma'$.  Let $K$ be the group consisting of all elements $k \in M$ such that both $\sigma(k)$ and $\sigma'(k)$ are scalar matrices.

Suppose $\sigma$ is not monomial.  From Lemma \ref{lem:P1P1versalMonomial} and Theorem \ref{thm:VersalCoxSplit}, we see that $K$ is non-trivial.  Suppose $K$ has an element $k$ of order $3$.  Either $k$ is central in $G$ or we may apply Lemma \ref{lem:A2toFermat} to see that $\rho$ is monomial.  In either case we have a contradiction.

It remains to consider $K$ a non-trivial abelian $2$-group.  Since $G$ has a trivial centre and $K$ is normal in $G$, there must be at least two distinct involutions in $K$.  Thus $K$ has rank at least $2$ and so there is some $k \in K$ such that $\rho(k)=(-\id_2,-\id_2)$.  This contradicts $G$ having a trivial centre.  Thus, $\sigma$ must be monomial and, so, $\rho$ is monomial.
\end{proof}

It remains only to prove the technical lemma used above:

\begin{lem}\label{lem:A2toFermat}
Suppose there exists an element $g \in G$ such that $g^3=1$ and $\rho(g)$ has $2$ distinct eigenvalues, each with multiplicity $2$.  Then $\rho$ is monomial.
\end{lem}

\begin{proof}
This theorem is essentially the $A_2$ case of \cite[Theorem 6.10]{DolgachevIskovskikh2006Finite-subgroup}.  Any surface with such a $g$ must be isomorphic to the ``Fermat cubic''.  All group actions on the Fermat cubic lift to monomial representations on $V$.  This proof is buried among other calculations so we sketch it here:

We see that $\rho(g)$ must have eigenvalues $\zeta_3^a$ and $\zeta_3^b$, with $a \ne b$, and the corresponding eigenspaces are $2$-dimensional.  Let us choose coordinates $\bbC[x_1,x_2,x_3,x_4]$ for the linear variety on which $g$ acts and let $\{x_1,x_2\}$ generate one eigenspace and $\{x_3,x_4\}$ the other.  When $g$ is considered as an element of $\PGL_4(\bbC)$, it suffices to assume $a=0$ and $b=1$.

Consider any monomial $x_1^{a_1}x_2^{a_2}x_3^{a_3}x_4^{a_4}$ in $F$.  Note that $g(F)=\zeta_3^{a_3+a_4}F$, so $a_3 + a_4$ is constant modulo $3$ for all monomials in $F$.  If $a_3 + a_4 \equiv 1 \mod 3$ then $x_3$, $x_4$ never occur with an exponent $\ge 2$.  Similarly, if $a_3 + a_4 \equiv 2 \mod 3$ then $x_1$, $x_2$ never occur with an exponent $\ge 2$.  Both of these are impossible by Lemma \ref{lem:minMonomialExp}.  The only possibility is $a_3 + a_4 \equiv 0 \mod 3$.  This means that all monomials in $F$ are of the form $x_1^{a_1}x_2^{a_2}$ or $x_3^{a_3}x_4^{a_4}$.

Thus, $F$ is of the form $F_1(x_1,x_2)=F_2(x_3,x_4)$ for homogeneous forms $F_1$ and $F_2$ of degree $3$.  The form $F_1$ does not have a multiple root at $(a:b)$ (as a point in $\bbP^1$) since otherwise $S$ would be singular at $(a:b:0:0)$.  We may choose new coordinates for $x_1,x_2$ taking the three roots to $(1:-1)$, $(1:-\zeta_3)$ and $(1:-\zeta_3^2)$ so that $F_1(x_1,x_2)=x_1^3+x_2^3$.  We may do the same for $F_2$, so we find $F$ is $a(x_1^3+x_2^3)+b(x_3^3+x_4^3)=0$ for $a,b \in \bbC^\times$.  Rescaling our variables again we may put $F$ in the form $x_1^3+x_2^3+x_3^3+x_4^3=0$.  This is the Fermat cubic.

Consider the subgroup $M \simeq C_3^4 \rtimes S_4$ of $\GL_4(\bbC)$ generated by permutating the variables and multiplying them by third roots of unity.  This group clearly preserves the form $x_1^3+x_2^3+x_3^3+x_4^3$.  The $27$ exceptional curves on $S$ are lines of the form $x_1+\eta x_2=x_3+\nu x_4=0$, $x_1+\eta x_3=x_2+\nu x_4=0$ or $x_1+\eta x_4=x_2+\nu x_3=0$ where $\eta^3=\nu^3=1$.  Any automorphism of $X$ must permute these lines.  $M$ contains all possible such automorphisms and is monomial.
\end{proof}

Thus all groups $G$ with versal actions on a del Pezzo surface of degree $3$ already act versally on $\bbP^2$, $\bbP^1 \times \bbP^1$ or $DP_5$.  This concludes the proof of Theorem \ref{thm:DP3Versal}.
\end{proof}